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 one year ago
If the period of a pendulum is 2.0 seconds and the acceleration due to gravity is 9.8 meters/second^2, What is the length of the cord on the pendulum?
A)1.0 meters
B)0.67 meters
C)1.3 meters
D)3.5 meters
 one year ago
If the period of a pendulum is 2.0 seconds and the acceleration due to gravity is 9.8 meters/second^2, What is the length of the cord on the pendulum? A)1.0 meters B)0.67 meters C)1.3 meters D)3.5 meters

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theEric
 one year ago
Best ResponseYou've already chosen the best response.1Are you speaking of a simple pendulum?

theEric
 one year ago
Best ResponseYou've already chosen the best response.1Then you'll want the equation that uses math to describe the relationships between different physical quantities. I think I've show it to you before. I found it at http://hyperphysics.phyastr.gsu.edu/hbase/pend.html It is:\[T=2\pi\sqrt{\frac{L}{g}}\] \(T\) is the period. \(L\) is the length of the chord. \(g\) is the acceleration caused by gravity. You want to solve for \(L\).

SNIIX18
 one year ago
Best ResponseYou've already chosen the best response.0I GOT O.67 IS THAT RIGHT

theEric
 one year ago
Best ResponseYou've already chosen the best response.1It's not what I got! How did you solve for \(L\)?

SNIIX18
 one year ago
Best ResponseYou've already chosen the best response.0I squared 2 then i divided 2.0 by 9.8

theEric
 one year ago
Best ResponseYou've already chosen the best response.1I'm confused... Can you show me what you got? You squared the 2?

SNIIX18
 one year ago
Best ResponseYou've already chosen the best response.0now im confused cause your confused

theEric
 one year ago
Best ResponseYou've already chosen the best response.1Haha, you start with \[T=2\pi\sqrt{\frac{L}{g}}\]Now, what did you square?

SNIIX18
 one year ago
Best ResponseYou've already chosen the best response.0owwww okay i see what i did wrong i was suppost to pi the 2 instead of square

theEric
 one year ago
Best ResponseYou've already chosen the best response.1I would divide the equation by \(2\pi\), square the equation, and multiply the equation by \(g\), if you know what I mean. Then you have \(L\) by itself after applicable cancelations, if you know what I mean.
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